To calculate the enthalpy of reaction ($\Delta H_{\text{rxn}}$) for the dissolution of NaOH, we will use the heat absorbed by the water, which can be calculated using the following formula:

$q = m \cdot C \cdot \Delta T$ Where:

• $q$ is the heat absorbed by the water,
• $m$ is the mass of the water,
• $C$ is the specific heat capacity of water (which is 4.18 J/g°C),
• $\Delta T$ is the change in temperature of the water.

Step 1: Determine the mass of the water

We are given that the volume of water is 50.00 mL, and we will assume the density of water is approximately 1.00 g/mL (since the temperature is near 20°C).

So, the mass of the water $m$ is:

$m_{\text{water}} = 50.00 \, \text{mL} \times 1.00 \, \text{g/mL} = 50.00 \, \text{g}$

Step 2: Calculate the heat absorbed by the water

We are also given that the initial temperature $T_{\text{initial}}$ is 20.1°C, and the final temperature $T_{\text{final}}$ is 28.6°C. Thus, the temperature change $\Delta T$ is:

$\Delta T = T_{\text{final}} - T_{\text{initial}} = 28.6°C - 20.1°C = 8.5°C$

Now, we can calculate the heat $q$ absorbed by the water:

$q = m_{\text{water}} \cdot C \cdot \Delta T$ $q = 50.00 \, \text{g} \times 4.18 \, \text{J/g°C} \times 8.5°C$ $q = 50.00 \times 4.18 \times 8.5 = 1776.5 \, \text{J}$

Step 3: Convert the heat to kJ

Since we want the result in kJ, convert the energy from joules (J) to kilojoules (kJ):

$q = 1776.5 \, \text{J} \times \frac{1 \, \text{kJ}}{1000 \, \text{J}} = 1.7765 \, \text{kJ}$

Step 4: Calculate the moles of NaOH dissolved

Next, we need to determine the number of moles of NaOH that dissolved. We are given the mass of NaOH, which is 4.01 g, and we can calculate the moles using the molar mass of NaOH.

The molar mass of NaOH is:

$\text{Molar mass of NaOH} = 22.99 \, (\text{Na}) + 15.999 \, (\text{O}) + 1.008 \, (\text{H}) = 39.997 \, \text{g/mol}$

Now, calculate the moles of NaOH:

$\text{moles of NaOH} = \frac{\text{mass of NaOH}}{\text{molar mass of NaOH}} = \frac{4.01 \, \text{g}}{39.997 \, \text{g/mol}} = 0.1003 \, \text{mol}$

Step 5: Calculate the enthalpy of reaction

The enthalpy of reaction $\Delta H_{\text{rxn}}$ is the heat per mole of NaOH dissolved. So, we can calculate it as:

$\Delta H_{\text{rxn}} = \frac{q}{\text{moles of NaOH}} = \frac{1.7765 \, \text{kJ}}{0.1003 \, \text{mol}} = 17.7 \, \text{kJ/mol}$

Thus, the enthalpy of reaction for the dissolution of NaOH is $\boxed{17.7} \, \text{kJ/mol}$.

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Would you like more details or have any further questions? Here are some related questions to explore:

1. How does the enthalpy of reaction change with different solutes?
2. What is the effect of temperature on the dissolution of NaOH?
3. How would the calculation change if the calorimeter had a different heat capacity?
4. How can we use Hess's Law to calculate enthalpy changes for other reactions?
5. What factors influence the solubility of NaOH in water?

Tip: To ensure accurate calculations when performing calorimetry experiments, always double-check the density and specific heat values, as they can vary with temperature.